#pressure at depth

Surface Tension, Capillarity and Excess Pressure

What is the force on a submarine window (0.06m2) at 400m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

Gauge pressure: Pg = ρgh = 1.03×103×9.8×400 = 4.0376×106Pa. F = PgA = 4.0376×106×0.06 = 2.42256×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.4 × 10⁵ N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Viscosity, Stokes' Law and Reynolds Number

What is the force on a submarine window (0.05m2) at 600m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

Gauge pressure: Pg = ρgh = 1.03×103×10×600 = 6.18×106Pa. F = PgA = 6.18×106×0.05 = 3.09×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 3.09 × 10⁵ N. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Bernoulli's Theorem and Applications

What is the total pressure at a depth of 4m in seawater (ρ\=1.03×103kg/m3) if atmospheric pressure is 1.01×105Pa? (Take

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.03×103kg/m3, g = 9.8m/s2, h = 4m. P = 1.01×105+1.03×103×9.8×4 = 1.01×105+40376 = 1.41376×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.41 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Viscosity, Stokes' Law and Reynolds Number

What is the total pressure at a depth of 1.6m in air (ρ\=1.29kg/m3) with atmospheric pressure 1.01×105Pa? (Take g\=9.8m/

Total pressure: P = Pa+ρgh. Pa = 1.01×105Pa, ρ = 1.29kg/m3, g = 9.8m/s2, h = 1.6m. P = 1.01×105+1.29×9.8×1.6 = 1.01×105+20.23 = 1.0102023×105Pa. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.01 × 10⁵ Pa. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Viscosity, Stokes' Law and Reynolds Number

What is the force on a submarine window (0.07m2) at 350m depth in seawater (ρ\=1.03×103kg/m3), interior at atmospheric p

Gauge pressure: Pg = ρgh = 1.03×103×10×350 = 3.605×106Pa. F = PgA = 3.605×106×0.07 = 2.5235×105N. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 2.5 × 10⁵ N. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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